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## Exercise 6A

**1. Add the following expressions:**

**(i) 5x, 7x, -6x**

**Solution:-**

In the above questions terms having the same literal factors are like terms.

Now add the like terms,

= 5x + 7x + (-6x) â€¦ [âˆµ + Ã— – = -]

Add terms having same sign first,

= 5x + 7x â€“ 6x

= 12x â€“ 6x

= 6x

**(ii) (3/5)x, (2/3)x, (-4/5)x**

**Solution:-**

In the above questions terms having the same literal factors are like terms.

= (3/5)x + (2/3)x + (-4/5)x

LCM of 5, 3, and 5 is 15

= (9x + 10x â€“ 12x)/ 15

= (19x â€“ 12x)/ 15

= (7/15)x

**(iii) 5a ^{2}b, -8a^{2}b, 7a^{2}b**

**Solution:-**

In the above questions terms having the same literal factors are like terms.

Now add the like terms,

= 5a^{2}b – 8a^{2}b + 7a^{2}b

= 12 a^{2}b – 8a^{2}b

= 4a^{2}b

**(iv) (3/4)x ^{2}, 5 x^{2}, -3x^{2}, -(1/4) x^{2}**

**Solution:-**

In the above questions terms having the same literal factors are like terms.

Now add the like terms,

= (3/4)x^{2} + 5x^{2} – 3x^{2} – (1/4) x^{2}

= (3/4)x^{2}– (1/4) x^{2} + 5x^{2} – 3x^{2}

= ((3-1)/4)x^{2} + 2x^{2}

= (2/4)x^{2} + 2x^{2}

= (1/2)x^{2} + 2x^{2}

= ((1 + 4)/2)x^{2}

= (5/2)x^{2}

**(v) x â€“ 3y + 4z, y â€“ 2x â€“ 8z, 5x â€“ 2y â€“ 3z**

**Solution:-**

Required sum,

= (x â€“ 3y + 4z) + (y â€“ 2x â€“ 8z) + (5x â€“ 2y â€“ 3z)

Collecting like terms,

= x â€“ 2x + 5x â€“ 3y + y â€“ 2y + 4z â€“ 8z â€“ 3z

Adding like terms,

= (1 â€“ 2 + 5)x + (- 3 + 1 â€“ 2)y + (4 â€“ 8 â€“ 3)z

= 4x â€“ 4y â€“ 7z

**(vi) 2x ^{2} â€“ 3y^{2}, 5x^{2} + 6y^{2}, -3x^{2} â€“ 4y^{2}**

**Solution:-**

Required sum,

= (2x^{2} â€“ 3y^{2}) + (5x^{2} + 6y^{2}) + (-3x^{2} â€“ 4y^{2})

Collecting like terms,

= 2x^{2} + 5x^{2} â€“ 3x^{2} â€“ 3y^{2} + 6y^{2} â€“ 4y^{2}

Adding like terms,

= (2 + 5 â€“ 3)x^{2} + (â€“ 3 + 6 â€“ 4)y^{2}

= 4x^{2} â€“ y^{2}

**(vii) 5x â€“ 2x ^{2} â€“ 8, 8x^{2 }â€“ 7x â€“ 9, 3 + 7x^{2} â€“ 2x**

**Solution:-**

Required sum,

= (5x â€“ 2x^{2} â€“ 8) + (8x^{2 }â€“ 7x â€“ 9) + (3 + 7x^{2} â€“ 2x)

Collecting like terms,

= 2x^{2} + 8x^{2} + 7x^{2} + 5x â€“ 7x â€“ 2x â€“ 8 â€“ 9 + 3

= (-2 + 8 + 7)x^{2} + (5 â€“ 7 â€“ 2)x + ( â€“ 8 â€“ 9 + 3)

= 13x^{2} â€“ 4x â€“ 14

**(viii) (2/3)a â€“ (4/5)v + (3/5)c, -(3/4)a â€“ (5/2)b + (2/3)c, (5/2)a + (7/4)b â€“ (5/6)c**

**Solution:-**

Required sum,

= [(2/3)a â€“ (4/5)b + (3/5)c] + [-(3/4)a â€“ (5/2)b + (2/3)c] + [(5/2)a + (7/4)b â€“ (5/6)c]

Collecting like terms,

= (2/3)a -(3/4)a + (5/2)a â€“ (4/5)b â€“ (5/2)b + (7/4)b + (3/5)c + (2/3)c â€“ (5/6)c

= [(2/3) – (3/4) + (5/2)]a + [(-4/5) â€“ (5/2) + (7/4)]b + [(3/5) + (2/3) â€“ (5/6)]c

= [(8 – 9 + 30)/12)a] + [(-16 -15 +30)/20)b] + [(18 + 20 â€“ 25)/ 30)c]

= (29/12)a +- (31/20)b + (13/30)c

**(ix) (8/5)x + (11/7)y + (9/4)xy, (-3/2)x â€“ (5/3)y â€“ (9/5)xy**

**Solution:-**

Required sum,

[(8/5)x + (11/7)y + (9/4)xy] + [(-3/2)x â€“ (5/3)y â€“ (9/5)xy]Collecting like terms,

= [(8/5)x – (3/2)x] + [(11/7)y â€“ (5/3)y] + [(9/4)xy â€“ (9/5)xy]

= (1/10)x â€“ (2/21)y + (9/20)xy

**(x) (3/2)x ^{3} – (1/4)x^{2} + (5/3), (-5/4)x^{3} + (3/5)x^{2} â€“ x + (1/5), -x^{2} + (3/8)x â€“ (8/15)**

**Solution:-**

Required sum,

= (3/2)x^{3} – (1/4)x^{2} + (5/3), (-5/4)x^{3} + (3/5)x^{2} â€“ x + (1/5), -x^{2} + (3/8)x â€“ (8/15)

Collecting like terms,

= [(3/2)x^{3} – (5/4)x^{3}] + [- (1/4)x^{2} + (3/5)x^{2} -x^{2}] + [â€“ x + (3/8)x] + [(5/3) + (1/5) â€“ (8/15)]

= (1/4)x^{3} â€“ (13/20)x^{2} â€“ (5/8)x + (4/3)

**2. Subtract :**

**(i) -8xy from 7xy**

**Solution:-**

The difference of two like terms is a like term whose coefficient is the difference of the numerical coefficient of the two like terms.

Then,

= (7 â€“ (-8))xy

= 15xy

**(ii) X ^{2} from -3x^{2}**

**Solution:-**

The difference of two like terms is a like term whose coefficient is the difference of the numerical coefficient of the two like terms.

Then,

= (-3 â€“ 1)x^{2}

= -4x^{2}

**(iii) (x â€“ y) from (4y â€“ 5x)**

**Solution:-**

The difference of two like terms is a like term whose coefficient is the difference of the numerical coefficient of the two like terms.

We have,

= (4y â€“ 5x) â€“ (x â€“ y)

Change the sign of each term of the expression to be subtracted and then add.

= 4y â€“ 5x â€“ x + y

= (-5x â€“ x) + (4y – y)

= -5x + 3y)

= 3y â€“ 5x

**(iv) (a ^{2} + b^{2} â€“ 2ab) from (a^{2} + b^{2} + 2ab)**

**Solution:-**

We have,

= (a^{2} + b^{2} + 2ab) – (a^{2} + b^{2} â€“ 2ab)

Change the sign of each term of the expression to be subtracted and then add.

= a^{2} + b^{2} + 2ab â€“ a^{2} â€“ b^{2} + 2ab

= (1 -1)a^{2} + (1 â€“ 1)b^{2} + (2 + 2)ab

= (0)a^{2} + (0)b^{2} + (4)ab

= 4ab

**(v) (x ^{2} â€“ y^{2}) from (2x^{2} â€“ 3y^{2} + 6xy)**

**Solution:-**

We have,

= (2x^{2} â€“ 3y^{2} + 6xy) – (x^{2} â€“ y^{2})

Change the sign of each term of the expression to be subtracted and then add.

= 2x^{2} â€“ 3y^{2} + 6xy â€“ x^{2} + y^{2}

= (2x^{2} – x^{2}) + (â€“ 3y^{2} + y^{2}) + 6xy

= (2 â€“ 1)x^{2} + (â€“ 3 + 1)y^{2} + 6xy

= 1x^{2} + (â€“ 2y^{2}) + 6xy

= 1x^{2} â€“ 2y^{2} + 6xy

**(vi) (x â€“ y + 3z) from (2z â€“ x – 3y)**

**Solution:-**

We have,

= (2z â€“ x – 3y) – (x â€“ y + 3z)

Change the sign of each term of the expression to be subtracted and then add.

= 2z â€“ x – 3y â€“ x + y â€“ 3z

= (2z â€“ 3z) + (â€“ x â€“ x) + (- 3y + y)

= (2 â€“ 3)z + (â€“ 1 â€“ 1)x + (- 3 +1)y

= -1z â€“ 2x – 2y

**3. Subtract (2a â€“ 3b + 4c) from the sum of (a + 3b â€“ 4c), (4a â€“ b + 9c) and (-2b + 3c â€“ a)**

**Solution:-**

First we find the sum of (a + 3b â€“ 4c), (4a â€“ b + 9c) and (-2b + 3c â€“ a)

= (a + 3b â€“ 4c) + (4a â€“ b + 9c) + (-2b + 3c â€“ a)

= (a + 3b â€“ 4c + 4a â€“ b + 9c – 2b + 3c â€“ a)

= (a + 4a â€“ a) + (3b â€“ b â€“ 2b) + (-4c + 9c + 3c)

= (1 + 4 â€“ 1)a + (3 â€“ 1 â€“ 2)b + (-4 + 9 + 3)c

= 4a + (0)b + 8c

=4a + 8c

Then,

Subtract (2a â€“ 3b + 4c) from (4a + 8c)

= (4a + 8c) – (2a â€“ 3b + 4c)

= 4a + 8c â€“ 2a + 3b â€“ 4c

= (4a â€“ 2a) + (3b) + (8c â€“ 4c)

= 2a + 3b + 4c

## Exercise 6B

**Find the products:**

**1. 3a ^{2} Ã— 8a^{4}**

**Solution:-**

The coefficient of the product of two monomials is equal to the product of their coefficients.

The variable part in the product of two monomials is equal to the product of the variables in the given monomials.

Then,

= (3 Ã— 8) Ã— (a^{2 }Ã— a^{4})

= (24) Ã— (a^{2+4}) â€¦ [âˆµ a^{m} Ã— a^{n} = a^{m+n}]

= (24) Ã— (a^{6})

= 24 a^{6+}

**2. -6x ^{3} Ã— 5x^{2}**

**Solution:-**

The coefficient of the product of two monomials is equal to the product of their coefficients.

The variable part in the product of two monomials is equal to the product of the variables in the given monomials.

Then,

= (-6 Ã— 5) Ã— (x^{3 }Ã— x^{2})

= (-30) Ã— (x^{3+2}) â€¦ [âˆµ a^{m} Ã— a^{n} = a^{m+n}]

= (-30) Ã— (x^{5})

= -30 a^{5}

**3.(-4ab) Ã— (-3a ^{2}bc)**

**Solution:-**

The coefficient of the product of two monomials is equal to the product of their coefficients.

The variable part in the product of two monomials is equal to the product of the variables in the given monomials.

Then,

= (-4 Ã— -3) Ã— (a Ã— a^{2}) Ã— (b Ã— b) Ã— c

= (12) Ã— (a^{1+2}) Ã— (b^{1+1}) Ã— c â€¦ [âˆµ a^{m} Ã— a^{n} = a^{m+n}]

= (12) Ã— (a^{3}) Ã— (b^{2}) Ã— c

= 12a^{3}b^{2}c

**4. (2a ^{2}b^{3}) Ã— (-3a^{3}b)**

**Solution:-**

The coefficient of the product of two monomials is equal to the product of their coefficients.

Then,

= (2 Ã— -3) Ã— (a^{2 }Ã— a^{3}) Ã— (b^{3} Ã— b)

= (-6) Ã— (a^{2+3}) Ã— (b^{3+1}) â€¦ [âˆµ a^{m} Ã— a^{n} = a^{m+n}]

= (-6) Ã— (a^{5}) Ã— (b^{4})

= -6a^{5}b^{4}

**5. (2/3)x ^{2}y Ã— (3/5)xy^{2}**

**Solution:-**

The coefficient of the product of two monomials is equal to the product of their coefficients.

Then,

= [(2/3) Ã— (3/5)] Ã— (x^{2 }Ã— x) Ã— (y Ã— y^{2})

= [(2Ã—3)/ (3Ã—5)] Ã— (x^{2+1}) Ã— (y^{1+2}) â€¦ [âˆµ a^{m} Ã— a^{n} = a^{m+n}]

= [(2Ã—1)/ (1Ã—5)] Ã— (x^{3}) Ã— (y^{3})

= [2/5]x^{3}y^{3}

**6. (-3/4)ab ^{3} Ã— (-2/3)a^{2}b^{4}**

**Solution:-**

The coefficient of the product of two monomials is equal to the product of their coefficients.

Then,

= [(-3/4) Ã— (-2/3)] Ã— (a Ã— a^{2}) Ã— (b^{3} Ã— b^{4})

= [(-3Ã—-2)/ (4Ã—3)] Ã— (a^{1+2}) Ã— (b^{3+4}) â€¦ [âˆµ a^{m} Ã— a^{n} = a^{m+n}]

= [(-1Ã—-1)/ (2Ã—1)] Ã— (a^{3}) Ã— (b^{7})

= [1/2]a^{3}b^{7}

**7. (-1/27)a ^{2}b^{2} Ã— (-9/2)a^{3}bc^{2}**

**Solution:-**

The coefficient of the product of two monomials is equal to the product of their coefficients.

Then,

= [(-1/27) Ã— (-9/2)] Ã— (a^{2 }Ã— a^{3}) Ã— (b^{2} Ã— b) Ã— c^{2}

= [(-1Ã—-9)/ (27Ã—2)] Ã— (a^{2+3}) Ã— (b^{2+1}) Ã— c^{2} â€¦ [âˆµ a^{m} Ã— a^{n} = a^{m+n}]

= [(-1Ã—-1)/ (3Ã—2)] Ã— (a^{5}) Ã— (b^{3}) Ã— c^{2}

= [1/6]a^{5}b^{3}c^{2}

**8. (-13/5)ab ^{2}c Ã— (7/3)a^{2}bc^{2}**

**Solution:-**

The coefficient of the product of two monomials is equal to the product of their coefficients.

Then,

= [(-13/5) Ã— (7/3)] Ã— (a Ã— a^{2}) Ã— (b^{2} Ã— b) Ã— (c Ã— c^{2})

= [(-13Ã—7)/ (5Ã—3)] Ã— (a^{1+2}) Ã— (b^{2+1}) Ã— (c^{1+2}) â€¦ [âˆµ a^{m} Ã— a^{n} = a^{m+n}]

= [-91/ 15] Ã— (a^{3}) Ã— (b^{3}) Ã— c^{3}

= [-91/15]a^{3}b^{3}c^{3}

**9. (-18/5)x ^{2}z Ã— (-25/6)xz^{2}y**

**Solution:-**

The coefficient of the product of two monomials is equal to the product of their coefficients.

Then,

= [(-18/5) Ã— (-25/6)] Ã— (x^{2 }Ã— x) Ã— (z Ã— z^{2}) Ã— (y)

= [(-18Ã—-25)/ (5Ã—6)] Ã— (x^{2+1}) Ã— (z^{1+2}) Ã— (y) â€¦ [âˆµ a^{m} Ã— a^{n} = a^{m+n}]

= [(-3Ã—-5)/ (1Ã—1)] Ã— (x^{3}) Ã— (z^{3}) Ã— y

= 15x^{3}z^{3}y

**10. (-3/14)xy ^{4} Ã— (7/6)x^{3}y**

**Solution:-**

The coefficient of the product of two monomials is equal to the product of their coefficients.

Then,

= [(-3/14) Ã— (7/6)] Ã— (x Ã— x^{3}) Ã— (y^{4} Ã— y)

= [(-3Ã—7)/ (14Ã—6)] Ã— (x^{1+3}) Ã— (y^{4+1}) â€¦ [âˆµ a^{m} Ã— a^{n} = a^{m+n}]

= [(-1Ã—1)/ (2Ã—2)] Ã— (x^{4}) Ã— (y^{5})

= (-1/4)x^{4}y^{5}

**11. (-7/5)x ^{2}y Ã— (3/2)xy^{2 } Ã— (-6/5)x^{3}y^{3}**

**Solution:-**

The coefficient of the product of two monomials is equal to the product of their coefficients.

Then,

= [(-7/5) Ã— (3/2) Ã— (-6/5)] Ã— (x^{2 }Ã— x Ã— x^{3}) Ã— (y Ã— y^{2} Ã— y^{3})

= [(-7Ã—3Ã—-6)/ (5Ã—2Ã—5)] Ã— (x^{2+1+3}) Ã— (y^{1+2+3}) â€¦ [âˆµ a^{m} Ã— a^{n} = a^{m+n}]

= [(-7Ã—3Ã—-2)/ (5Ã—1Ã—5)] Ã— (x^{6}) Ã— (y^{6})

= (63/25)x^{6}y^{6}

**12. 2a ^{2}b Ã— (-5)ab^{2}c Ã— (-6)bc^{2}**

**Solution:-**

The coefficient of the product of two monomials is equal to the product of their coefficients.

Then,

= [(2) Ã— (-5) Ã— (-6)] Ã— (a^{2 }Ã— a) Ã— (b Ã— b^{2 }Ã— b) Ã— (c Ã— c^{2})

= [60] Ã— (a^{2+1}) Ã— (b^{1+2+1}) Ã— (c^{1+2}) â€¦ [âˆµ a^{m} Ã— a^{n} = a^{m+n}]

= [60] Ã— (a^{3}) Ã— (b^{4}) Ã— (c^{3})

= [60]a^{3}b^{4}c^{3}

## Exercise 6C

**Find each of the following products:**

**1. 4a(3a + 7b)**

**Solution**:-

Let p, q and r be three monomials.

Then, by distributive law of multiplication over addition, we have:

P Ã— (q + r) = (p Ã— q) + (p Ã— r)

Now,

= (4a Ã— 3a) + (4a Ã— 7b)

= (12a^{2} + 28ab) â€¦ [âˆµ a^{m} Ã— a^{n} = a^{m+n}]

**2. 5a(6a – 3b)**

**Solution**:-

Let p, q and r be three monomials.

Then, by distributive law of multiplication over subtraction, we have:

P Ã— (q – r) = (p Ã— q) – (p Ã— r)

Now,

= (5a Ã— 6a) – (5a Ã— 3b)

= (30a^{2} – 15ab) â€¦ [âˆµ a^{m} Ã— a^{n} = a^{m+n}]

**3. 8a ^{2}(2a + 5b)**

**Solution**:-

Let p, q and r be three monomials.

Then, by distributive law of multiplication over addition, we have:

P Ã— (q + r) = (p Ã— q) + (p Ã— r)

Now,

= (8a^{2} Ã— 2a) + (8a^{2} Ã— 5b)

= (16a^{3} + 40a^{2}b) â€¦ [âˆµ a^{m} Ã— a^{n} = a^{m+n}]

**4. 9x ^{2}(5x + 7)**

**Solution**:-

Let p, q and r be three monomials.

Then, by distributive law of multiplication over addition, we have:

P Ã— (q + r) = (p Ã— q) + (p Ã— r)

Now,

= (9x^{2} Ã— 5x) + (9x^{2} Ã— 7)

= (45x^{3} + 63x^{2}) â€¦ [âˆµ a^{m} Ã— a^{n} = a^{m+n}]

**5. ab(a ^{2} â€“ b^{2})**

**Solution**:-

Let p, q and r be three monomials.

Then, by distributive law of multiplication over subtraction, we have:

P Ã— (q – r) = (p Ã— q) – (p Ã— r)

Now,

= (ab Ã— a^{2}) – (ab Ã— b^{2})

= (a^{3}b + ab^{3}) â€¦ [âˆµ a^{m} Ã— a^{n} = a^{m+n}]

**6. 2x ^{2}(3x â€“ 4x^{2})**

**Solution**:-

Let p, q and r be three monomials.

Then, by distributive law of multiplication over subtraction, we have:

P Ã— (q – r) = (p Ã— q) – (p Ã— r)

Now,

= (2x^{2} Ã— 3x) – (2x^{2} Ã— 4x^{2})

= (6x^{3} â€“ 8x^{4}) â€¦ [âˆµ a^{m} Ã— a^{n} = a^{m+n}]

**7.(3/5)m ^{2}n(m + 5n)**

**Solution**:-

Let p, q and r be three monomials.

Then, by distributive law of multiplication over addition, we have:

P Ã— (q + r) = (p Ã— q) + (p Ã— r)

Now,

= ((3/5)m^{2}n Ã— m) + ((3/5)m^{2}n Ã— 5n)

= ((3/5)m^{3}n + 3m^{2}n^{2}) â€¦ [âˆµ a^{m} Ã— a^{n} = a^{m+n}]

**8. -17x ^{2}(3x â€“ 4)**

**Solution**:-

Let p, q and r be three monomials.

Then, by distributive law of multiplication over subtraction, we have:

P Ã— (q – r) = (p Ã— q) – (p Ã— r)

Now,

= (-17x^{2} Ã— 3x) – (-17x^{2} Ã— 4)

= (-51x^{3} + 68x^{2}) â€¦ [âˆµ a^{m} Ã— a^{n} = a^{m+n}]

**9. (7/2)x ^{2}((4/7)x + 2)**

**Solution**:-

Let p, q and r be three monomials.

Then, by distributive law of multiplication over addition, we have:

P Ã— (q + r) = (p Ã— q) + (p Ã— r)

Now,

= ((7/2)x^{2} Ã— (4/7)x) + ((7/2)x^{2} Ã— 2)

= ((7Ã—4)/ (2Ã—7))x^{3} + ((7Ã—2)/(2Ã—1))x^{2} â€¦ [âˆµ a^{m} Ã— a^{n} = a^{m+n}]

= ((1Ã—2)/ (1Ã—1))x^{3} + ((7Ã—1)/(1Ã—1))x^{2}

= (2)x^{3} + (7)x^{2}

**10. -4x ^{2}y(3x^{2} â€“ 5y)**

**Solution**:-

Let p, q and r be three monomials.

Then, by distributive law of multiplication over subtraction, we have:

P Ã— (q – r) = (p Ã— q) – (p Ã— r)

Now,

= (-4x^{2}y Ã— 3x^{2}) – (-4x^{2}y Ã— -5y)

= (-12x^{4}y + 20x^{2}y^{2}) â€¦ [âˆµ a^{m} Ã— a^{n} = a^{m+n}]

**11. (-4/27)xyz((9/2)x ^{2}yz â€“ (3/4)xyz^{2})**

**Solution**:-

Let p, q and r be three monomials.

Then, by distributive law of multiplication over subtraction, we have:

P Ã— (q – r) = (p Ã— q) – (p Ã— r)

Now,

= ((-4/27)xyz Ã— (9/2)x^{2}yz) – ((-4/27)xyz Ã— (-3/4)xyz^{2})

= ((-4Ã—9)/ (27Ã—2))x^{3}y^{2}z^{2} + ((-4Ã—-3)/(27Ã—4))x^{2}y^{2}z^{3} â€¦ [âˆµ a^{m} Ã— a^{n} = a^{m+n}]

= ((-2Ã—1)/ (3Ã—1))x^{3}y^{2}z^{2} + ((-1Ã—-1)/(9Ã—1))x^{2}y^{2}z^{3}

= (-2/3)x^{3}y^{2}z^{2} + (1/9)x^{2}y^{2}z^{3}

**12. 9t ^{2}(t + 7t^{3})**

**Solution**:-

Let p, q and r be three monomials.

Then, by distributive law of multiplication over addition, we have:

P Ã— (q + r) = (p Ã— q) + (p Ã— r)

Now,

= (9t^{2} Ã— t) + (9t^{2} Ã— 7t^{3})

= (9t^{3} + 63t^{5}) â€¦ [âˆµ a^{m} Ã— a^{n} = a^{m+n}]

**13. 10a ^{2}(0.1a â€“ 0.5b)**

**Solution**:-

Let p, q and r be three monomials.

Then, by distributive law of multiplication over subtraction, we have:

P Ã— (q – r) = (p Ã— q) – (p Ã— r)

Now,

= (10a^{2} Ã— 0.1a) – (10a^{2} Ã— 0.5b)

= (1a^{3} â€“ 5a^{2}b) â€¦ [âˆµ a^{m} Ã— a^{n} = a^{m+n}]

## Exercise 6D

**Find each of the following products:**

**1. (5x + 7) (3x + 4)**

**Solution:-**

Suppose (a + b) and (c + d) are two binomials. By using the distributive law of multiplication over addition twice, we may find their product as given below.

(a + b) Ã— (c + d) = a Ã— (c + d) + b Ã— (c + d) = (a Ã— c + a Ã— d) + (b Ã— c + b Ã— d)

= ac + ad + bc + bd

Let,

a= 5x, b= 7, c= 3x, d= 4

Now,

= 5x Ã— (3x + 4) +7 Ã— (3x + 4)

= [(5x Ã— 3x) + (5x Ã— 4)] + [(7 Ã— 3x) + (7 Ã— 4)]

= [15x^{2} + 20x + 21x + 28]

= [15x^{2} + 41x + 28]

**2. (4x – 3) (2x + 5)**

**Solution:-**

Suppose (a – b) and (c + d) are two binomials. By using the distributive law of multiplication over addition twice, we may find their product as given below.

(a – b) Ã— (c + d) = a Ã— (c + d) – b Ã— (c + d) = (a Ã— c + a Ã— d) – (b Ã— c + b Ã— d)

= ac + ad – bc – bd

Let,

a= 4x, b= 3, c= 2x, d= 5

Now,

= 4x Ã— (2x + 5) -3 Ã— (2x + 5)

= [(4x Ã— 2x) + (4x Ã— 5)] – [(3 Ã— 2x) + (3 Ã— 5)]

= [8x^{2} + 20x – 6x – 15]

= [8x^{2} + 14x – 15]

**3. (x – 6) (4x + 9)**

**Solution:-**

Suppose (a – b) and (c + d) are two binomials. By using the distributive law of multiplication over addition twice, we may find their product as given below.

(a – b) Ã— (c + d) = a Ã— (c + d) – b Ã— (c + d) = (a Ã— c + a Ã— d) – (b Ã— c + b Ã— d)

= ac + ad – bc – bd

Let,

a= x, b= 6, c= 24x, d= 9

Now,

= x Ã— (4x + 9) -6 Ã— (4x + 5)

= [(x Ã— 4x) + (x Ã— 9)] – [(6 Ã— 4x) + (6 Ã— 9)]

= [4x^{2} + 9x – 24x – 54]

= [4x^{2} – 15x – 54]

**4. (5y – 1) (3y – 8)**

**Solution:-**

Suppose (a – b) and (c – d) are two binomials. By using the distributive law of multiplication over addition twice, we may find their product as given below.

(a – b) Ã— (c – d) = a Ã— (c – d) – b Ã— (c – d) = (a Ã— c – a Ã— d) – (b Ã— c – b Ã— d)

= ac – ad – bc + bd

Let,

a= 5y, b= 1, c= 3y, d= 8

Now,

= 5y Ã— (3y – 8) -1 Ã— (3y – 8)

= [(5y Ã— 3y) + (5y Ã— -8)] – [(1 Ã— 3y) + (1 Ã— -8)]

= [15y^{2} â€“ 40y â€“ 3y + 8]

= [15y^{2} â€“ 43y + 8]

**5. (7x + 2y) (x + 4y)**

**Solution:-**

Suppose (a + b) and (c + d) are two binomials. By using the distributive law of multiplication over addition twice, we may find their product as given below.

(a + b) Ã— (c + d) = a Ã— (c + d) + b Ã— (c + d) = (a Ã— c + a Ã— d) + (b Ã— c + b Ã— d)

= ac + ad + bc + bd

Let,

a= 7x, b= 2y, c= x, d= 4y

Now,

= 7x Ã— (x + 4y) +2y Ã— (x + 4y)

= [(7x Ã— x) + (7x Ã— 4y)] + [(2y Ã— x) + (2y Ã— 4y)]

= [7x^{2} + 28xy + 2yx + 8y^{2}]

= [7x^{2} + 30xy + 8y^{2}]

**6. (9x + 5y) (4x + 3y)**

**Solution:-**

Suppose (a + b) and (c + d) are two binomials. By using the distributive law of multiplication over addition twice, we may find their product as given below.

(a + b) Ã— (c + d) = a Ã— (c + d) + b Ã— (c + d) = (a Ã— c + a Ã— d) + (b Ã— c + b Ã— d)

= ac + ad + bc + bd

Let,

a= 9x, b= 5y, c= 4x, d= 3y

Now,

= 9x Ã— (4x + 3y) +5y Ã— (4x + 3y)

= [(9x Ã— 4x) + (9x Ã— 3y)] + [(5y Ã— 4x) + (5y Ã— 3y)]

= [36x^{2} + 27xy + 20yx + 15y^{2}]

= [36x^{2} + 47xy + 15y^{2}]

**7. (3m â€“ 4n) (2m â€“ 3n)**

**Solution:-**

Suppose (a – b) and (c – d) are two binomials. By using the distributive law of multiplication over addition twice, we may find their product as given below.

(a – b) Ã— (c – d) = a Ã— (c – d) – b Ã— (c – d) = (a Ã— c – a Ã— d) – (b Ã— c – b Ã— d)

= ac – ad – bc + bd

Let,

a= 3m, b= 4n, c= 2m, d= 3n

Now,

= 3m Ã— (2m â€“ 3n) -4n Ã— (2m â€“ 3n)

= [(3m Ã— 2m) + (3m Ã— -3n)] – [(4n Ã— 2m) + (4n Ã— -3n)]

= [6m^{2} â€“ 9mn â€“ 8mn + 12n^{2}]

= [6m^{2} â€“ 17mn + 12n^{2}]

**8. (0.8x â€“ 0.5y) (1.5x â€“ 3y)**

**Solution:-**

Suppose (a – b) and (c – d) are two binomials. By using the distributive law of multiplication over addition twice, we may find their product as given below.

(a – b) Ã— (c – d) = a Ã— (c – d) – b Ã— (c – d) = (a Ã— c – a Ã— d) – (b Ã— c – b Ã— d)

= ac – ad – bc + bd

Let,

a= 0.8x, b= 0.5y, c= 1.5x, d= 3y

Now,

= 0.8x Ã— (1.5x â€“ 3y) – 0.5y Ã— (1.5x â€“ 3y)

= [(0.8x Ã— 1.5x) + (0.8x Ã— -3y)] – [(0.5y Ã— 1.5x) + (0.5y Ã— -3y)]

= [1.2x^{2} â€“ 2.4xy â€“ 0.75yx + 1.5y^{2}]

= [1.2x^{2} â€“ 3.15xy + 1.5y^{2}]

**9. ((1/5)x + 2y) ((2/3)x â€“ y)**

**Solution:-**

Suppose (a + b) and (c – d) are two binomials. By using the distributive law of multiplication over addition twice, we may find their product as given below.

(a + b) Ã— (c – d) = a Ã— (c – d) + b Ã— (c – d) = (a Ã— c – a Ã— d) + (b Ã— c – b Ã— d)

= ac – ad + bc – bd

Let,

a= (1/5)x, b= 2y, c= (2/3)x, d= y

Now,

= (1/5)x Ã— ((2/3)x â€“ y) + 2y Ã— ((2/3)x â€“ y)

= [((1/5)x Ã— (2/3)x) + ((1/5)x Ã— -y)] + [(2y Ã— (2/3)x) + (2y Ã— -y)]

= [(2/15)x^{2} – (1/5)xy + (4/3)yx – 2y^{2})]

= [(2/15)x^{2} + (17/15) xy – 2y^{2}]

**10. ((2/5)x â€“ (1/2)y) (10x â€“ 8y)**

**Solution:-**

(a – b) Ã— (c – d) = a Ã— (c – d) – b Ã— (c – d) = (a Ã— c – a Ã— d) – (b Ã— c – b Ã— d)

= ac – ad – bc + bd

Let,

a= (2/5)x, b=(1/2)y, c= 10x, d= 8y

Now,

= (2/5)x Ã— (10x â€“ 8y) – (1/2)y Ã— (10x â€“ 8y)

= [((2/5)x Ã— 10x) + ((2/5)x Ã— -8y)] – [((1/2)y Ã— 10x) + ((1/2)y Ã— -8y)]

= [4x^{2} â€“ (16/5)xy â€“ 5yx + 4y^{2}]

= [4x^{2} â€“ (41/5)xy + 4y^{2}]

**11. ((3/4)a + (2/3)b) (4a + 3b)**

**Solution:-**

(a + b) Ã— (c + d) = a Ã— (c + d) + b Ã— (c + d) = (a Ã— c + a Ã— d) + (b Ã— c + b Ã— d)

= ac + ad + bc + bd

Let,

a= (3/4)a, b=(2/3)b, c= 4a, d= 3b

Now,

= (3/4)a Ã— (4a + 3b) + (2/3)b Ã— (4a + 3b)

= [((3/4)a Ã— 4a) + ((3/4)a + 3b)] – [((2/3)b Ã— 4a) + ((2/3)b Ã— + 3b)]

= [3a^{2} + (9/4)ab + (8/3)ab + 2b^{2}]

= [3a^{2} + ((27+32)/12)ab + 2b^{2}]

= [3a^{2} + (59/12)ab + 2b^{2}]

**12. (x ^{2} â€“ a^{2}) (x â€“ a)**

**Solution:-**

(a – b) Ã— (c – d) = a Ã— (c – d) – b Ã— (c – d) = (a Ã— c – a Ã— d) – (b Ã— c – b Ã— d)

= ac – ad – bc + bd

Let,

a= x^{2}, b= a^{2}, c= x, d= a

Now,

= x^{2} Ã— (x â€“ a) – a^{2} Ã— (x â€“ a)

= [(x^{2} Ã— x) + (x^{2} Ã— -a)] – [(a^{2} Ã— x) + (a^{2} Ã— -a)]

= [x^{3} â€“ x^{2}a â€“ a^{2}x + a^{3}]

**13. (3p ^{2} + q^{2}) (2p^{2} â€“ 3q^{2})**

**Solution:-**

Suppose (a + b) and (c – d) are two binomials. By using the distributive law of multiplication over addition twice, we may find their product as given below.

(a + b) Ã— (c – d) = a Ã— (c – d) + b Ã— (c – d) = (a Ã— c – a Ã— d) + (b Ã— c – b Ã— d)

= ac – ad + bc – bd

Let,

a= 3p^{2}, b= q^{2}, c= 2p^{2}, d= 3q^{2}

Now,

= 3p^{2}Ã— (2p^{2} â€“ 3q^{2}) + q^{2} Ã— (2p^{2} â€“ 3q^{2})

= [(3p^{2}Ã— 2p^{2}) + (3p^{2}Ã— -3q^{2})] + [(q^{2} Ã— 2p^{2}) + (q^{2} Ã— -3q^{2})]

= [6p^{4} â€“ 9p^{2}q^{2} + 2q^{2}p^{2} â€“ 3q^{4})]

= [6p^{4} â€“ 7p^{2}q^{2} â€“ 3q^{4}]

**14. (2x ^{2} â€“ 5y^{2}) (x^{2} + 3y^{2})**

**Solution:-**

Suppose (a – b) and (c + d) are two binomials. By using the distributive law of multiplication over addition twice, we may find their product as given below.

(a – b) Ã— (c + d) = a Ã— (c + d) – b Ã— (c + d) = (a Ã— c + a Ã— d) – (b Ã— c + b Ã— d)

= ac + ad – bc – bd

Let,

a= 2x^{2}, b= 5y^{2}, c= x^{2}, d= 3y^{2}

Now,

= 2x^{2} Ã— (x^{2} + 3y^{2}) â€“ 5y^{2} Ã— (x^{2} + 3y^{2})

= [(2x^{2} Ã— x^{2}) + (2x^{2} Ã— 3y^{2})] – [(5y^{2} Ã— x^{2}) + (5y^{2} Ã— 3y^{2})]

= [2x^{4} + 6x^{2}y^{2} â€“ 5y^{2}x^{2} â€“ 15y^{4}]

= [2x^{4} + x^{2}y^{2} â€“ 15y^{4}]

**15. (x ^{3} â€“ y^{3}) (x^{2} + y^{2})**

**Solution:-**

(a – b) Ã— (c + d) = a Ã— (c + d) – b Ã— (c + d) = (a Ã— c + a Ã— d) – (b Ã— c + b Ã— d)

= ac + ad – bc – bd

Let,

a= x^{3}, b= y^{3}, c= x^{2}, d= y^{2}

Now,

= x^{3} Ã— (x^{2} + y^{2}) â€“ y^{3} Ã— (x^{2} + y^{2})

= [(x^{3} Ã— x^{2}) + (x^{3} Ã— y^{2})] – [(y^{3} Ã— x^{2}) + (y^{3} Ã— y^{2})]

= [x^{5} + x^{3}y^{2} â€“ y^{3}x^{2} â€“ y^{5}]

## RS Aggarwal Solutions for Class 7 Maths Chapter 6 – Algebraic Expressions

Chapter 6 – Algebraic Expressions contains 4 exercises and the RS Aggarwal Solutions available on this page provide solutions to the questions present in the exercises. Now, let us have a look at some of the concepts discussed in this chapter.

- Constants and Variables
- Addition of Algebraic Expressions
- Subtraction of Algebraic Expressions
- Multiplication of Algebraic Expressions
- Multiplication of Monomials
- Multiplication of a Monomial and a Binomial
- Multiplication of Two Binomials

### Chapter Brief of RS Aggarwal Solutions for Class 7 Maths Chapter 6 – Algebraic Expressions

RS Aggarwal Solutions for Class 7 Maths Chapter 6 – Algebraic Expressions. A symbol having a fixed numeric value is called a constant and a symbol which takes various numerical values is called variables. A combination of constants and variables connected by the signs of addition, subtraction, multiplication and division is called algebraic expressions. In this chapter, we shall learn about simple algebraic expressions involving one or two variables. We shall also study their addition and subtraction.